spoj QTREE3 Query on a tree again!

You are given a tree (an acyclic undirected connected graph) with N
nodes. The tree nodes are numbered from 1 to N. In the start, the
color of any node in the tree is white.

We will ask you to perfrom some instructions of the following form:

0 i : change the color of the i-th node (from white to black, or from black to white);or1 v : ask for the id of the first black node on the path from node 1 to node v. if it doesn't exist, you may return -1 as its result. 

Input

In the first line there are two integers N and Q.

In the next N-1 lines describe the edges in the tree: a line with two
integers a b denotes an edge between a and b.

The next Q lines contain instructions “0 i” or “1 v” (1 ≤ i, v ≤ N).
Output

For each “1 v” operation, write one integer representing its result.

树链剖分，维护路径上深度最小的黑点。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int fir[100010],ne[200010],to[200010],val[1000010],dep[100010],fa[100010],size[100010],son[100010],pos[100010],top[100010],n,q,clo;void add(int num,int u,int v){    ne[num]=fir[u];    fir[u]=num;    to[num]=v;}void dfs1(int u){    int v;    size[u]=1;    for (int i=fir[u];i;i=ne[i])        if ((v=to[i])!=fa[u])        {            dep[v]=dep[u]+1;            fa[v]=u;            dfs1(v);            size[u]+=size[v];            if (!son[u]||size[son[u]]<size[v]) son[u]=v;        }}void dfs2(int u){    int v;    pos[u]=++clo;    if (son[u])    {        top[son[u]]=top[u];        dfs2(son[u]);    }    for (int i=fir[u];i;i=ne[i])        if ((v=to[i])!=fa[u]&&v!=son[u])        {            top[v]=v;            dfs2(v);        }}void build(int p,int L,int R){    val[p]=-1;    if (L==R) return;    int mid=L+R>>1;    if (L<=mid) build(p*2,L,mid);    if (mid<R) build(p*2+1,mid+1,R);}void init(){    int u,v;    scanf("%d%d",&n,&q);    for (int i=1;i<n;i++)    {        scanf("%d%d",&u,&v);        add(i*2,u,v);        add(i*2+1,v,u);    }    dep[1]=1;    dfs1(1);    top[1]=1;    dfs2(1);    build(1,1,n);}void modi(int p,int L,int R,int u){    int x,y;    if (L==R)    {        if (val[p]==-1) val[p]=u;        else val[p]=-1;        return;    }    int mid=L+R>>1;    if (pos[u]<=mid) modi(p*2,L,mid,u);    else modi(p*2+1,mid+1,R,u);    x=L<=mid?val[p*2]:-1;    y=mid<R?val[p*2+1]:-1;    if (x==-1||(y>0&&dep[y]<dep[x])) val[p]=y;    else val[p]=x;}int qry(int p,int L,int R,int l,int r){    int x,y;    if (l<=L&&R<=r) return val[p];    int mid=L+R>>1;    x=l<=mid?qry(p*2,L,mid,l,r):-1;    y=r>mid?qry(p*2+1,mid+1,R,l,r):-1;    if (x==-1||(y>0&&dep[y]<dep[x])) return y;    return x;}int query(int u){    int ans=-1,x;    while (top[u]!=1)    {        x=qry(1,1,n,pos[top[u]],pos[u]);        if (x>0&&(ans==-1||dep[x]<dep[ans])) ans=x;        u=fa[top[u]];    }    x=qry(1,1,n,1,pos[u]);    if (x>0&&(ans==-1||dep[x]<dep[ans])) ans=x;    return ans;}int main(){    int u,x;    init();    while (q--)    {        scanf("%d%d",&x,&u);        if (x) printf("%d\n",query(u));        else modi(1,1,n,u);    }}